Problem: In square $ABCD$, $AD$ is 4 centimeters, and $M$ is the midpoint of $\overline{CD}$. Let $O$ be the intersection of $\overline{AC}$ and $\overline{BM}$. What is the ratio of $OC$ to $OA$? Express your answer as a common fraction.

[asy]

size (3cm,3cm);

pair A,B,C,D,M;

D=(0,0);
C=(1,0);
B=(1,1);
A=(0,1);

draw(A--B--C--D--A);

M=(1/2)*D+(1/2)*C;

draw(B--M);

draw(A--C);

label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$O$",(0.5,0.3));
label("$M$",M,S);

[/asy]
Explanation: First we draw diagonal $\overline{BD}$, and let the diagonals intersect at $T$, as shown:


[asy]

size (4cm,4cm);

pair A,B,C,D,M;

D=(0,0);
C=(1,0);
B=(1,1);
A=(0,1);

draw(A--B--C--D--A);

M=(1/2)*D+(1/2)*C;

draw(B--M);

draw(A--C);

label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$O$",(0.5,0.3));
label("$M$",M,S);
draw(B--D);
label("$T$",(B+D)/2,N);
[/asy]

Since $\overline{CT}$ and $\overline{BM}$ are medians of $\triangle BCD$, point $O$ is the centroid of $\triangle BCD$, so $OC= (2/3)CT$.  Since $T$ is the midpoint of $\overline{AC}$, we have $CT = AC/2$, so $OC= (2/3)CT = (2/3)(AC/2) = AC/3$.  Since $\overline{OC}$ is $\frac13$ of $\overline{AC}$, we know that $\overline{OA}$ is the other $\frac23$ of $\overline{AC}$, which means $OC/OA = \boxed{\frac{1}{2}}$.